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JEE Advanced 2016 Paper 2 · Chemistry · Atomic Structure · medium

For the hydrogen-like ion He^+ (Z=2), take the Bohr radius a0=52.9 pm, the hydrogen ground-state energy -13.6 eV and the Rydberg constant…

One or more options correct

For the hydrogen-like ion He+\mathrm{He^+} (Z=2Z=2), take the Bohr radius a0=52.9a_0=52.9 pm, the hydrogen ground-state energy −13.6-13.6 eV and the Rydberg constant RH=1.097×107 m−1R_H=1.097\times10^{7}\ \mathrm{m^{-1}}. Select the correct statement(s).
  1. A. The radius of the n=2n=2 orbit of He+\mathrm{He^+} is about 106106 pm.
  2. B. The energy of the n=3n=3 level of He+\mathrm{He^+} is −13.6-13.6 eV.
  3. C. The longest-wavelength line of the Lyman-type series of He+\mathrm{He^+} (n→1n\to1) has λ≈30.4\lambda\approx30.4 nm.
  4. D. The energy needed to remove the electron of He+\mathrm{He^+} from its n=2n=2 level is 13.613.6 eV.
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Answer: A, C, D

For a hydrogen-like species rn=a0n2/Zr_n=a_0n^2/Z and En=−13.6 Z2/n2E_n=-13.6\,Z^2/n^2 eV. (A) r2=52.9×4/2=105.8r_2=52.9\times4/2=105.8 pm: correct. (B) E3=−13.6×4/9=−6.04E_3=-13.6\times4/9=-6.04 eV, whereas −13.6-13.6 eV would be the value for n=2n=2 of He+\mathrm{He^+} (or n=1n=1 of H): incorrect. (C) The longest wavelength of the n→1n\to1 series is for n=2→1n=2\to1: 1/λ=RHZ2(1−14)=3RH1/\lambda=R_HZ^2(1-\tfrac14)=3R_H, so λ=1/(3×1.097×107)=3.04×10−8\lambda=1/(3\times1.097\times10^7)=3.04\times10^{-8} m =30.4=30.4 nm: correct. (D) The energy needed to remove the electron from n=2n=2 is ∣E2∣=13.6×4/4=13.6|E_2|=13.6\times4/4=13.6 eV: correct. The Z2Z^2 scaling thus makes He+\mathrm{He^+} (n=2n=2) energetically equal to H (n=1n=1). Correct options: A, C and D.
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Original question written in the JEE Advanced 2016 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake